What is the de Sitter temperature as of now? It is claimed that de Sitter temperature is $$T=\frac{1}{2\pi}H,$$
where $H$ is the Hubble constant. I presume it is expressed in natural units with which I am not familiar. So what it will be in Kelvins? Is it higher than the temperature of CMB? If so, is there any heat exchange between the both?
 A: I give the answer with the general method even though a much straightforward way would be to guess the result because it is simple.
The unit of $H$ is the inverse of time denoted by $[\mathrm T^{-1}]$. The dimension of a temperature is denoted by $[\Theta]$. To find the numerical value of $T$ in kelvin, one should find a combination of $c:[\mathrm{LT^{-1}}]$, $\hbar=[\mathrm{ML^2T^{-1}}]$, $\mathcal G:[\mathrm{M^{-1}L^3T^{-2}}]$ and $k_{\mathrm B}:[\mathrm{ML^2T^{-2}\Theta^{-1}}]$ ($[\mathrm L]$ and $[\mathrm M]$ represent length and mass respectively) such that
$$ c^x \;\hbar^y \;\mathcal G ^z\; k_{\mathrm B}^u\times H$$
has the dimension of a temperature ($x$, $y$, $z$ and $u$ are unknown). 
This gives the system
$$\left\{\begin{array}{rcc}
y-z+u&=0&\quad[\mathrm M]\\
x+2y+3z+2u&=0&\quad[\mathrm L]\\
-x-y-2z-2u&=1&\quad[\mathrm T]\\
-u&=1&\quad[\Theta]
\end{array}\right.$$
The solution is
$$\left\{\begin{array}{cl}
x&=0\\
y&=1\\
z&=0\\
u&=-1
\end{array}\right.$$
We obtain thus
$$T=\frac{\hbar}{2\pi k_{\mathrm B}}H.$$
The value of $H$ is $H=67.8\, \mathrm{km.s^{-1}.Mpc^{-1}}=2.194\times 10^{-18}\,\mathrm{m.s^{-1}}$. We find a temperature of $T=2.67\times10^{-30}\,\mathrm K$. 
